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Laurent · #2 December 29th, 2008, 08:55 AM

Quote:

Originally Posted by geton

The amount of vegetables eaten by a family in a week is a random variable W kg. The probability density function is given by:

$f(w) = \begin{cases} \frac {20}{5^5}w^3(5-w), &\text {$0 \leq w \leq 5$,} \\ 0, & \text {otherwise.} \end{cases}$

(a) Find the cumulative distribution function of W.
(b) Find the probability that the family eats between 2kg & 4kg of vegetables in one week.
(c) Verify that the amount, m, of vegetables such that the family is equally likely to eat more of less than m in any week is about 3.431 kg.

I have done part (a) & (b), and the answer of (b) is 0.650. But how could I verify (c)?

I agree with your answer to (b). As for (c), you are asked to check that $P(W\leq m)\simeq P(W\geq m)$ for $m=3.421\,{\rm kg}$ . Since the two probabilities sum to 1 (justify), this amounts to checking that $P(W\leq m)\simeq\frac{1}{2}$ (I wrote $\simeq$ since $m$ is "about" 3.421 kg). You have to compute an integral, like in (b).